>Yes. But nowhere is either the numerator *or* the denominator 0 until the>limit is reached. (I'll repeat what I just said in reply to another post...>I neglected to mention that k is positive.) Therefore (I hope...I agree this >is the weakness in the proof), since substituting k for x yields equivalent >functions, the limit of the quotient would be 1.>

But this is definitely not true. You seem to want to think thatif f and g are continuous functions,then lim f(x)/g(x) = f(k)/g(k) x->kbut it isn't true unless f(x)/g(x) is itself continuous at k,i.e. unless g(k) not equal to 0.

it's not clear what you mean when you say"substituing k for x yields equivalent functions" -- once you substitute k for x, you no longer have functions, you have specific numbers.

as someone else pointed out, choosing any k for which sin(k^2) = 0would work in your argument, whether or not k was an integer.